Erase-Remove Idiom

Lab C++ Intermediate 20 points

Read n integers and remove all negative numbers using remove_if + erase. Print the remaining numbers (or empty).

Input

Line 1: n. Line 2: n integers.

Output

One line.

Example

Input

6
3 -1 4 -1 -5 9

Output

3 4 9

Sample input available Sample output available

Editor

C++

Hints

remove_if moves the kept items to the front and returns the new logical end.

erase then deletes the leftover tail.

Solution explanation

This two-step erase-remove idiom is the standard way to delete by condition before C++20.

Show solution

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (int &x : v) cin >> x;
    v.erase(remove_if(v.begin(), v.end(), [](int x) { return x < 0; }), v.end());
    if (v.empty()) cout << "empty";
    for (size_t i = 0; i < v.size(); i++) cout << v[i] << (i + 1 < v.size() ? " " : "");
    cout << "\n";
    return 0;
}

Output

Answer